How prove this nice limit $\lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}f(\{ka\})=\int_{0}^{1}f(x)dx$

let $f(x)$ is Continuous on $[0,1]$, and such $f(0)=f(1)$,and if $a$ is irrational number.

show that

$$\lim_{n\to+\infty}\dfrac{1}{n}\sum_{k=1}^{n}f(\{ka\})=\int_{0}^{1}f(x)dx$$

where $\{ka\}=x-[x]$,and $[x]$ is is the largest integer not greater than $x$

This problem is from this ( problem 8)http://wenku.baidu.com/view/a643e6c26137ee06eff91855.html

and I find this Prove that $\lim_{N\rightarrow\infty}(1/N)\sum_{n=1}^N f(nx)=\int_{0}^1f(t)dt$ Have without Trigonometric series methods? because this problem is Freshman exam questions

-
If one really wants Freshman Calculus level solution, one might have to reconstruct the proof of Stone-Weierstrass Theorem from scratch. –  i707107 Oct 7 '13 at 18:24
Also, the solution in your link is wrong –  i707107 Oct 7 '13 at 18:25
This is essentially part of the Weyl's criterion, which requires some amount of analysis stuffs. –  sos440 Oct 7 '13 at 19:15
Hell0,@Sos440,can you post your solution? Thank you –  user94270 Oct 8 '13 at 2:17
@i707107,why this link solution is wrong? –  user94270 Oct 8 '13 at 2:18

I doubt that there is a trigonometry-free proof of the equidistribution property (see statement below). Here, I will deduce the result in the OP from this equidistribution property (because the solution proposed in the linked answer is wrong, as has been pointed out).

Equidistribution property Let $I=[b,c]$ be a subinterval of $[0,1]$. Then the frequency $f_n(I)=\frac{w_n(I)}{n}$ where $w_n(I)$ is the number of $k\in[n]=\lbrace 1,2,3, \ldots, n\rbrace$ such that $\lbrace k\alpha \rbrace \in I$ satisfies $f_n(I) \to {\sf length}(I)=c-b$ when $n\to +\infty$.

Solving your exercise with the equidistribution property

Let $\varepsilon > 0$. Let $\eta$ be another positive number, depending on $\varepsilon$, to be defined later. Since $|f|$ is continuous on the compact set $[0,1]$, it reaches a maximum there, which we will denote by $M$. Since $f$ is continuous on the compact set $[0,1]$, it is also uniformly continuous there. So there is a $\delta >0$ such that for any $x,y\in [0,1]$

$$|x-y| \leq \delta \Rightarrow |f(x)-f(y)| \leq \eta \tag{1}$$

Let $q$ be an integer such that $\frac{1}{2^q} \leq \delta$. Then the intervals $I_j=[\frac{j-1}{2^q},\frac{j}{2^q}] (1 \leq j \leq 2^q)$ make a subdivision of $[0,1]$. By the equidistribution property, for each $j$ the sequence $(f_n(I_j))_{n\geq 1}$ tends to $\frac{1}{2^q}$ when $n\to +\infty$. So there is an integer $n_0(q,\eta)$ such that for any $n \geq n_0(q,\eta)$ and $1\leq j \leq 2^q$,

$$\bigg|f_n(I_j)-\frac{1}{2^q}\bigg| \leq \eta \tag{2}$$

Next, define

$$X_{n,j}=\bigg\lbrace k\in [n] \ \bigg| \ \lbrace ka \rbrace \in I_j \bigg\rbrace \tag{3}$$

If we put $d_n=\frac{\sum_{k=1}^{n} f(\lbrace ka\rbrace)}{n}-\int_{[0,1]}f$, then we have $d_n=\displaystyle\sum_{k=1}^{2^q} d_{n,j}$ where

$$d_{n,j}=\frac{\displaystyle\sum_{k\in X_{n,j}}f(\lbrace ka \rbrace)}{n}- \int_{I_j} f \tag{4}$$

On each interval $I_j$, $f$ is continuous and so attains a minimum value $m_j$. We then have

$$\begin{array}{lclc} |d_{n,j}| & \leq & \Bigg| {\displaystyle\sum_{k\in X_{n,j}}\frac{f(\lbrace ka \rbrace)-m_j}{n}} \Bigg|+ \Bigg| {\displaystyle\sum_{k\in X_{n,j}}\frac{m_j}{n}-\int_{I_j}m_j} \Bigg|+ \Bigg| \int_{I_j}(f-m_j) \Bigg| \\ & = & \Bigg| {\displaystyle\sum_{k\in X_{n,j}}\frac{f(\lbrace ka \rbrace)-m_j}{n}} \Bigg|+ |m_j|\Bigg|f_n(I_j)-\frac{1}{2^q}\Bigg|+ \Bigg| \int_{I_j}(f-m_j) \Bigg| \\ & \leq & {\displaystyle\sum_{k\in X_{n,j}}\frac{|f(\lbrace ka \rbrace)-m_j|}{n}} +M\eta+ \int_{I_j}|f-m_j| \\ & \leq & {\displaystyle\sum_{k\in X_{n,j}}\frac{\eta}{n}} +M\eta+ \int_{I_j}\eta \\ & \leq & \eta +M\eta+ \frac{\eta}{2^q}=\eta(M+1+\frac{1}{2^q}) \\ \end{array}$$

Summing on $j$, we deduce

$$|d_n| \leq \eta (2^q(M+1)+1)$$

Taking $\eta=\frac{\varepsilon}{2^q(M+1)+1}$, we have $|d_n| \leq \varepsilon$, so we have shown that $(d_n)$ tends to zero as wished.

-

Let see “Concrete mathematics : a foundation for computer science” book, chapter 3, section 3.5, pages 87–89.

-

There actually is a proof not using trigonometric series but heavily using ergodic theory. I guess that wouldn't be a freshman's solution.

Anyhow, the major idea behind the ergodic theory proof is as follows: One identifies $[0,1]$ with the unit circle $S^1 = \{e^{2\pi i t}\mid t \in \mathbb{R}\}$. Consider on $S^1$ the rotation $R$ with angle $2\pi\alpha$: $R(e^{2\pi i t}) = e^{2\pi i (t+\alpha)}$. The sequence $(\{k\alpha\})_k$ translates on $S^1$ to $(R^k(1))_k$. Now one endows $S^1$ with its Borel $\sigma$-algebra and shows that $R$ is uniquely ergodic with the Haar measure (corresponding to the Lebesgue measure on $[0,1]$) being the unique ergodic $R$-invariant measure. The statement then follows from Birkhoff's pointwise ergodic theorem.

More details are contained in many textbooks on ergodic theory. This precise example is worked out in, e.g., Einsiedler, Ward: Ergodic Theory with a view towards Number Theory.

-